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嗨，希望通过HP3458A数字化1 kHz（1 Vpp）正弦波，并将采样数据传输到主机pc。我已经下载了示例程序，4100点程序，在c＃中。
我可以让程序运行，创建样本并返回数据。它使用DCV数字化模式。现在，我想优化设置，使其特定于1kHz正弦波。示例测试程序代码：cmds =“PRESET DIG; “; //预设到指定状态（数字化）cmds + =“MFORMAT DINT;”; //设置内存存储格式（双整数）cmds + =“OFORMAT ASCII;”; //设置输出格式（ASCII）cmds + =“MEM FIFO;”; //清除内存并设置内存存储类型cmds + =“APER 2E-6;”; //设置arperature时间（cmds + =“TRIG AUTO;”; //设置触发源（Auto）cmds + =“NRDGS 4100，AUTO;”; //设置读数（4100读数）cmds + = “END ON;”; //启用EOI功能cmds + =“TARM HOLD”; //保持触发直到触发SendCmd（“TARM SGL”）;我的程序，试图将1kHz正弦数字化：cmds =“PRESET DIG; “; cmds + =”MFORMAT DINT;“; cmds + =”OFORMAT ASCII;“; cmds + =”MEM FIFO;“; cmds + =”APER 0.0000122;“; cmds + =”tiMER 0.0000122;“; cmds + = “NRDGS 2048，TIMER;”; cmds + =“END ON;”; cmds + =“TARM HOLD”; SendCmd（“TARM SGL”）;返回TARM SGL。2048（0到2047）样本的程序错误 然而，最后两个样本都没有了。当我完成一个FFT时，基本位于bin 51.通过上述设置，我希望这是1kHz，而bin #25的基波是我的采样数据。 相信我的设置应该代表这个数字化样本集：F0 = 1000.576 HzM = 25 CyclesN = 2048 SmplsFs = 81967.21311 HzFres = 40.023 HzI刚刚关闭 1kHz使Timer和Arp值保持100 nS的倍数。对于什么是错误的任何想法。 以上来自于谷歌翻译 以下为原文 Hi, Looking to digitize a 1 kHz (1 Vpp) sine wave via the HP3458A, and transfer the sampled data to the host pc. I've downloaded the example program, 4100 point program, in c#. I can get the program to function, create samples and return data. It uses the DCV digitizing mode. Now, i'd like to optimize the settings so it's specific to a 1kHz sinewave. Example test program code: cmds = "PRESET DIG;"; // Preset to the designated state (Digitizing) cmds += "MFORMAT DINT;"; // Set the memory storage format (double integer) cmds += "OFORMAT ASCII;"; // Set the output format (ASCII) cmds += "MEM FIFO;"; // Clear memory and set memory storage type cmds += "APER 2E-6;"; // Set the arperature time ( cmds += "TRIG AUTO;"; // Set the trigger source (Auto) cmds += "NRDGS 4100,AUTO;"; // Set the number of readings (4100 readings) cmds += "END ON;"; // Enable EOI function cmds += "TARM HOLD"; // Hold the trigger until triggered SendCmd("TARM SGL"); My Program, trying to digitize a 1kHz sine: cmds = "PRESET DIG;"; cmds += "MFORMAT DINT;"; cmds += "OFORMAT ASCII;"; cmds += "MEM FIFO;"; cmds += "APER 0.0000122;"; cmds += "TIMER 0.0000122;"; cmds += "NRDGS 2048,TIMER;"; cmds += "END ON;"; cmds += "TARM HOLD"; SendCmd("TARM SGL"); My program errors duing the TARM SGL. 2048 (0 to 2047) samples are returned, however the last two samples are out of wack. When I complete an FFT, the fundamental is at bin 51. With the above setting, i expect this to be 1kHz and a fundamental at bin# 25. Attached is my sampled data. I believe my setting should represent this digitizing sample set: F0 = 1000.576 Hz M = 25 Cycles N = 2048 Smpls Fs = 81967.21311 Hz Fres = 40.023 Hz I chose just off 1kHz to keep the Timer and Arp value a multipal of 100 nS. Any ideas as to what's wrong. 
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5个回答
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我玩弄了这个，发现你错过了TRIG AUTO命令，并且我得到了一个“触发太快”的错误，除非我放慢了计时器的速度。
请参阅下面的我的更改。我没有进行FFT，但我输入了一个2 Vpp 1 kHz正弦波，具有-1 V偏移，数据看起来很好。我编程，试图将1kHz正弦数字化：cmds =“PRESET DIG ;“; cmds + =“MFORMAT DINT;”; cmds + =“OFORMAT ASCII;”; cmds + =“MEM FIFO;”; cmds + =“APER 0.0000122;”; “TRIG AUTO”“TIMER 0.00003”cmds + =“NRDGS 2048，TIMER;”; cmds + =“END ON;”; cmds + =“TARM HOLD”; SendCmd（“TARM SGL”）; 以上来自于谷歌翻译 以下为原文 I played around with this and found that you were missing the TRIG AUTO command and that I got a “trigger too fast” error unless I slowed down the timer. See my changes below. I didn’t do an FFT, but I input a 2 Vpp 1 kHz sine wave with a -1 V offset and the data looks good. My Program, trying to digitize a 1kHz sine: cmds = "PRESET DIG;"; cmds += "MFORMAT DINT;"; cmds += "OFORMAT ASCII;"; cmds += "MEM FIFO;"; cmds += "APER 0.0000122;"; “TRIG AUTO” “TIMER 0.00003” cmds += "NRDGS 2048,TIMER;"; cmds += "END ON;"; cmds += "TARM HOLD"; SendCmd("TARM SGL"); |
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谢谢你的快速回复。哦，你的两个更新修复了我得到的错误。但我现在正在努力解决Aperture时间，样本数量和TIMER时间的关联。我的理解：孔径时间= 1 /采样率= 0.0000122
秒秒Fs = SamplingRate = 1 / TIMER = 81967.2131147539 Hz N =采样数= 2048光圈和TIMER时间只能以100 nS的增量进行更改，然后略微偏离1kHz的音调。 如果采样率高于以上且输入音调为Fo = 1000.576 Hz，（Fs / N = Fo / M），则预期M = 25个周期。 捕获的数据有比这更多的周期。 知道了，在FFT之后，我希望基波在bin＃25中，因为Fres将是40.023 Hz。你可以帮助我将孔径时间，TIMER时间和样本数量等于拟合相干采样方程（Fs / N = Fo / M）。 我知道我不会连贯，因为3458与我的源不同步，但涂抹的基础应该散布在正确的bin中。 一个窗口稍后会帮助我。欣赏你的时间！ 附上新的正弦数据。 以上来自于谷歌翻译 以下为原文 Thanks for the quick reply. Ok, your two updates fix's the error's I was getting. But I am now struggling with the association of the Aperture time, Number of Samples and TIMER time. My understanding: Aperture time = 1/Sample Rate = 0.0000122 Seconds Fs = SamplingRate = 1/TIMER = 81967.2131147539 Hz N = Number Of Samples = 2048 Aperture and TIMER time can only be changed in 100 nS increments, hece the slightly off 1kHz tone. With a SamplingRate of Above and Input tone of Fo = 1000.576 Hz, (Fs / N = Fo / M) i would expect M = 25 cycles. Data captured has many more cycles than that. Knowing that, post FFT, i would expect the fundamental to be in bin# 25 as the Fres would be 40.023 Hz. Can you help me equate aperture time, TIMER time and number of samples to fit the Coherent sampling equation (Fs / N = Fo / M). I know I will not be coherent as the 3458 is not synchronized to my source, but the smeared fundamental should be spread around the correct bin. A window will later help me out there. Appreciate your time! Attached new sine data. 附件
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至少部分答案是3458A具有集成A / D而不是像示波器那样的闪存A / D.
光圈时间是A / D对输入信号进行积分的实际时间。 采样率是触发A / D时的速率，这是= 1 / TIMER。 触发和测量开始之间可能存在延迟，以便在必要时进行稳定。 请参阅3458A用户指南的第160页：指定值FYIIf如果需要超过50KHz，则必须使用MFORMAT SINT和OFORMAT SINT 以上来自于谷歌翻译 以下为原文 At least part of the answer is that the 3458A has an integrating A/D not a flash A/D like a scope. Aperture time is the actual time the A/D is integrating the input signal. The sample rate is the rate at with the A/D is triggered and this does = 1/TIMER. There can be a delay between the trigger and the measurement start to allow settling if necessary. See page 160 of the 3458A User’s Guide: Specifying a value <500ns selects minimum aperture which is 500ns. FYI If you need to go faster than 50KHz you must use MFORMAT SINT and OFORMAT SINT |
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